Problem: Solve for $x$. Enter the solutions from least to greatest. Round to two decimal places. $(x + 3)^2 - 3 = 0$ $\text{lesser }x = $
Solution: $\begin{aligned} (x + 3)^2 - 3&= 0 \\\\ (x+3)^2&=3 \\\\ \sqrt{(x+3)^2}&=\sqrt{3} \end{aligned}$ $\begin{aligned} x+3&=\pm\sqrt 3 \\\\ x&=\pm\sqrt 3-3 \\ \phantom{(x + 3)^2 - 3}& \\ x=-\sqrt 3-3&\text{ or }x=\sqrt3-3 \\\\ x\approx -4.73&\text{ or }x\approx -1.27 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -4.73 \\\\ \text{greater } x &= -1.27 \end{aligned}$